1. Show that the function f : R∗ → R∗
defined by f(x) = 1/x is one-one and onto,
where R∗
is the set of all non-zero real numbers. Is the result true, if the domain
R∗
is replaced by N with co-domain being same as R∗?

Answer

2. Check the injectivity and surjectivity of the following functions:

(i) f : N → N given by f(x) = x2

(ii) f : Z → Z given by f(x) = x2

(iii) f : R → R given by f(x) = x2

(iv) f : N → N given by f(x) = x3

(v) f : Z → Z given by f(x) = x3

Answer

3. Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither
one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Answer

f : R → R is given by, f(x) = [x]

It is seen that f(1.2) = [1.2] = 1

f(1.9)= [1.9] = 1

∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9

∴ f is not one-one.

Now, consider 0.7 ∈ R.

It is known that f(x) = [x] is always an integer.

Thus, there does not exist any element x ∈ R such that f(x) = 0.7.

∴ f is not onto.

Hence, the greatest integer function is neither one-one nor onto.

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4. Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither one-ne
nor onto, where |x| is x, if x is positive or 0 and |x| is – x, if x is negative.

Answer

It is clear that f(-1) = I-1| = 1

and f(1) = I1l = 1

∴ f(-1) = f(1), but -1 ≠ 1

∴ f is not one-one.

Now, consider -1 ∈ R.

It is known that f(x) = Ix| is always non-negative.

Thus, there does not exist any element x in domain R such that f(x) = Ix| = -1.

∴ f is not onto.

Hence, the modulus function is neither one-one nor onto.

5. Show that the Signum Function f : R → R, given by

Answer

It is seen that f(1) = f(2) = 1, but 1 ≠ 2.

∴ f is not one-one.

Now, as f(x) takes only 3 values (1, 0, or -1) for the element -2 in co-domain R, there does not exist any x in domain R such that f(x) = -2.